AQA A-Level Chemistry Paper 3, 2019: Question 27

✅ Correct Answer

D: [(C₂H₅)₂N(CH₃)₂]⁺ Br⁻

The starting amine is methylamine (CH₃NH₂), which possesses only one methyl group. The electrophile being added in each step is an ethyl group (C₂H₅–) from bromoethane. Therefore, any product derived from this mixture can only contain one methyl group attached to the nitrogen atom. Product D has two methyl groups, making it impossible to form here.

💡 Key Knowledge

• Nucleophile vs Electrophile: The amine acts as a nucleophile via the lone pair on nitrogen; the haloalkane carbon (δ⁺) is attacked.
• Further Alkylation: Because the product secondary and tertiary amines still possess a lone pair on the nitrogen atom, they remain nucleophilic and react with further molecules of haloalkane.
• Tracking the Alkyl Groups:
  • Initial nucleophile: CH₃NH₂ (1 methyl)
  • Reagent added: C₂H₅Br (adds ethyl groups)
  • Every successive product retains exactly 1 methyl group and gains ethyl groups.

📐 Step-by-Step Reaction Pathway

1. Initial Substitution (forms a secondary amine):
   CH₃NH₂ + C₂H₅Br → C₂H₅NHCH₃ + HBr
   → Matches Option A (ethylmethylamine)
2. Second Substitution (forms a tertiary amine):
   C₂H₅NHCH₃ + C₂H₅Br → (C₂H₅)₂NCH₃ + HBr
   → Matches Option B (diethylmethylamine)
3. Third Substitution (forms a quaternary salt):
   (C₂H₅)₂NCH₃ + C₂H₅Br → [(C₂H₅)₃NCH₃]⁺ Br⁻
   → Matches Option C (triethylmethylammonium bromide)

Notice how every valid product contains exactly one –CH₃ group and an increasing number of –C₂H₅ groups.

🧠 Exam Technique

• Read the negative: The question asks which compound is NOT a possible product. Highlight or circle the word "not" immediately.
• Count carbon groups first: In multiple-alkylation questions, do not waste time writing out full mechanisms. Simply count:
  • How many methyl groups are present in the starting amine? → 1
  • What group is being introduced? → Ethyl
• Scan the options for any compound containing more than one methyl group — D contains (CH₃)₂ and is instantly identifiable in under 15 seconds.

❌ Common Errors & Pitfalls

• Confusing the reactants: Assuming bromoethane and methylamine could swap roles and somehow form dimethyl derivatives. For a second methyl group to attach, bromomethane (CH₃Br) would need to be present.
• Misinterpreting quaternary salts: Some students assume quaternary salts cannot form because all N–H bonds are gone. Remember, tertiary amines still have a lone pair to form a dative covalent bond with a haloalkane to give a quaternary ammonium salt (Option C is possible!).
• Rushing the bracket subscripts: Misreading [(C₂H₅)₃NCH₃]⁺ vs [(C₂H₅)₂N(CH₃)₂]⁺ under exam pressure. Always verify the numbers outside brackets carefully.